AQA A-Level Chemistry Paper 2, 2021: Question 7

9 marks · Medium difficulty · State/Explain/Describe

Spectroscopy and Structure Determination This question explores the use of spectroscopy to identify organic compounds. Part (a) involves interpreting an infrared (IR) spectrum of a compound with molecular formula C₄H₈O, where students must identify the functional group. In part (b), another compound (C₄H₇NO) is reduced, and students must describe how the IR spectrum changes. Part (c) tests interpretation of proton NMR data for C₃H₇ClO, where students analyse chemical shifts, integration values, and splitting patterns to deduce the structure.

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Question

AQA A-Level Chemistry Paper 2, 2021: Question 7
Question text

07 This question is about spectroscopy.

07.1 Compound K has molecular formula C4H8O

Figure 5 shows the infrared spectrum of K.

Figure 5

Which functional group does K contain?

Tick ( ) one box.

[1 mark]

Functional Group

alcohol alkene amine carbonyl nitrile

07.2 Compound L has molecular formula C4H7NO

Figure 6 shows the infrared spectrum of L.

Figure 6

L reacts with H2 in the presence of a nickel catalyst to give compound M.

*19* Suggest three ways in which the infrared spectrum of M is different from the infrared

spectrum of L.

[3 marks]

07.3 Figure 7 shows the 1H NMR spectrum of Q, C H ClO

Figure 7

Table 4 shows the chemical shifts (δ values) and integration values for each peak.

Table 4

δ value / ppm 3.95 3.65 3.35

Integration value 0.6 0.6 0.9

Deduce the structure of Q.

Explain your answer.

[5 marks]

Mark scheme

Show the mark scheme Mark scheme for AQA A-Level Chemistry Paper 2, 2021: Question 7

Question Answers Additional Comments/Guidelines Mark

07.1 Tick in carbonyl box only 1

Peak at 2220-2260 cm-1 (for C≡N) disappears If both C≡N disappears and N-H appears without M1

wavenumbers scores 1

07.2 Peak at 3300-3500 cm-1 (for N-H) appears M2

Fingerprint region different M3

Integration ratio 2:2:3 If no link between delta value and oxygen and M1

chlorine, then can award 1 mark for correct

Peak at 3.95 triplet (integration 2) Cl-CH2 next to CH2 explanation of splitting of all 3 peaks M2

07.3 Peak at 3.65 triplet (integration 2) O-CH2 next to CH2 If no explanation of splitting, then can award 1 mark M3

for 3 correct links between delta value and oxygen

Peak at 3.35 singlet (integration 3) O-CH no adjacent H and chlorine M4

M5

Structure CH3-O-CH2CH2Cl

How to answer it

Study Guide: Spectroscopy – IR and 1H NMR

Q1. Functional Group from IR Spectrum

Compound K (C₄H₈O) shows a strong peak just below 1750 cm–1.

This indicates: Carbonyl group (C=O)

Carbonyls absorb between 1680–1750 cm–1. No broad O–H peak (alcohols show one at 3200–3550 cm–1) → not an alcohol.
Don’t confuse C=O with C=C (around 1620 cm–1) or O–H stretches.

Q2. IR Changes When C=N Is Reduced

Compound L → M (via H₂/Ni reduction)

Changes in IR spectrum:

  1. Peak for C=N at ~2220–2260 cm–1 disappears
  2. New peak for N–H at ~3300–3500 cm–1 appears
  3. Fingerprint region (600–1500 cm–1) will also change
Look for functional group peaks in specific regions — C=N is sharp and strong; N–H is broader like O–H but slightly lower.

Q3. Deduce the Structure of Q (NMR)

Molecular formula: C₃H₇ClO → must be fully saturated (no C=C or rings)

Integration values: 0.6 : 0.6 : 0.9 → equivalent to 2:2:3

  • 3.95 ppm triplet (2H): CH2 next to Cl → ClCH₂–
  • 3.65 ppm triplet (2H): CH2 next to O → –CH₂O–
  • 3.35 ppm singlet (3H): CH₃ not next to H → –OCH₃

Structure: CH₃OCH₂CH₂Cl

Triplets mean –CH₂– groups are adjacent to one other hydrogen-bearing carbon. The singlet suggests no adjacent H atoms.
Less than 20% scored 3+ marks here — splitting patterns and linking to neighbours was a key weakness.

Topics

Organic Chemistry · 3.3.6 Organic Analysis · 3.3.15 Nuclear Magnetic Resonance Spectroscopy

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.